What is the general solution for tan^5(x)=tan(x) ?

The general solution for tan^5(x)=tan(x) is x=pin,x=(3pi)/4+pin,x= pi/4+pin

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tan^5(x)=tan(x)

Solution

tan^{5} (x) = tan (x)

Solution

x = πn, x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{5} (x) = tan (x)

Solve by substitution

tan^{5} (x) = tan (x)

Let:

tan (x) = u

u^{5} = u

u^{5} = u : u = 0, u = i, u = - i, u = - 1, u = 1

u^{5} = u

Move

u

to the left side

u^{5} = u

Subtract

u

from both sides

u^{5} - u = u - u

Simplify

u^{5} - u = 0

u^{5} - u = 0

Factor

u^{5} - u : u (u^{2} + 1) (u + 1) (u - 1)

u^{5} - u

Factor out common term

u : u (u^{4} - 1)

u^{5} - u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{5} = u^{4} u

= u^{4} u - u

Factor out common term

u

= u (u^{4} - 1)

= u (u^{4} - 1)

Factor

u^{4} - 1 : (u^{2} + 1) (u + 1) (u - 1)

u^{4} - 1

Rewrite

u^{4} - 1

(u^{2})^{2} - 1^{2}

u^{4} - 1

Rewrite

1

1^{2}

= u^{4} - 1^{2}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

u^{4} = (u^{2})^{2}

= (u^{2})^{2} - 1^{2}

= (u^{2})^{2} - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(u^{2})^{2} - 1^{2} = (u^{2} + 1) (u^{2} - 1)

= (u^{2} + 1) (u^{2} - 1)

Factor

u^{2} - 1 : (u + 1) (u - 1)

u^{2} - 1

Rewrite

1

1^{2}

= u^{2} - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

u^{2} - 1^{2} = (u + 1) (u - 1)

= (u + 1) (u - 1)

= (u^{2} + 1) (u + 1) (u - 1)

= u (u^{2} + 1) (u + 1) (u - 1)

u (u^{2} + 1) (u + 1) (u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or u^{2} + 1 = 0 or u + 1 = 0 or u - 1 = 0

Solve

u^{2} + 1 = 0 : u = i, u = - i

u^{2} + 1 = 0

Move

1

to the right side

u^{2} + 1 = 0

Subtract

1

from both sides

u^{2} + 1 - 1 = 0 - 1

Simplify

u^{2} = - 1

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

The solutions are

u = 0, u = i, u = - i, u = - 1, u = 1

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = i, tan (x) = - i, tan (x) = - 1, tan (x) = 1

tan (x) = 0, tan (x) = i, tan (x) = - i, tan (x) = - 1, tan (x) = 1

tan (x) = 0 : x = πn

tan (x) = 0

General solutions for

tan (x) = 0

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan (x) = i :

No Solution

tan (x) = i

No Solution

tan (x) = - i :

No Solution

tan (x) = - i

No Solution

tan (x) = - 1 : x = \frac{3 π}{4} + πn

tan (x) = - 1

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = πn, x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan^5(x)=tan(x) ?
The general solution for tan^5(x)=tan(x) is x=pin,x=(3pi)/4+pin,x= pi/4+pin